# java recursive solution

• Repeatedly select 2 numbers (all combinations) and compute, until there is only one number, check if it's 24.

``````class Solution {
public boolean judgePoint24(int[] nums) {
return f(new double[] {nums[0], nums[1], nums[2], nums[3]});
}

private boolean f(double[] a) {
if (a.length == 1) {
return a[0] == 24;
}
for (int i = 0; i < a.length; i++) {
for (int j = i + 1; j < a.length; j++) {
double[] b = new double[a.length - 1];
for (int k = 0, l = 0; k < a.length; k++) {
if (k != i && k != j) {
b[l++] = a[k];
}
}
for (double k : compute(a[i], a[j])) {
b[a.length - 2] = k;
if (f(b)) {
return true;
}
}
}
}
return false;
}

private double[] compute(double a, double b) {
return new double[] {a + b, a - b, b - a, a * b, a / b, b / a};
}
}
``````

• @2499370956

I think you need to handle the divide-by-zero case.

• I think "double" data type have a solution for divide-by-zero case. For the reference: https://stackoverflow.com/questions/12954193/why-does-division-by-zero-with-floating-point-or-double-precision-numbers-not

• @jun1013
cool. Did not know that before. thx.

• Great solution! However, I think they added some other test cases (if this solution passed all test cases before). Try [3,3,8,8], what you supposed to get is "true" (why? Tricky like this: 8 / (3 - (8 / 3)) = 24), but this solution will yield "false". I think this may have to do with double data type (lost some significant number during the computation?) . I modified your solution by changing `return a[0] == 24` to `if (Math.abs(a[0] - 24) < 0.0001) return true` . Now this will pass all the test cases as of 9/18/2017.

• @vonzcy Any time.

• @2499370956
Hi,
Your solution is so good. Actually better than most solutions(Like some straightforward brute force solution) I can find by now for me. However it cannot pass all the test cases caused by this command "return a[0] == 24". The problem is when a number is not divisible, you wont get exactly "24". Actually you may use an "err" to compare two double - type numbers. Like, double err = 0.0001, return Math.abs(a[0] - 24) < err.
Thanks.

• @Candle309 right

• @jun1013 Thx for your help. But if I wrote this as return Math.abs(a[0] - 24) < 0.0001
it would go wrong, do you know why?

• @2499370956 how do you handle divided by 0 here?
thanks

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